Question

Find the sum of the first $n$ terms of the series :$3+7+13+21+31+...$

Answer 1

The given series is $S=3+7+13+21+31+....$

$\Rightarrow S=3+7+13+21+31+...a_{n-1}+a_n$

$\Rightarrow S=3+7+13+21+31+...a_{n-2}+a_{n-1}+a_n$

On subtracting both the equation, we obtain

$S-S=[3+(7+13+21+31+...+a_{n-1}+a_n\left)\right]-\left[\right(3+7+13+21+31+...+a_{n-1}+a_n]$

$\Rightarrow S-S=3+\left[\right(7-3)+(13-7)+(21-13)+...(a_n+a_{n-1}\left)\right]a_n$

$\Rightarrow 0=3+[4+6+8+...(n-1\left)terms\right]-a_n$

$\Rightarrow a_n=3+[4+6+8+...(n-1\left)terms\right]$

$\Rightarrow a_n=3+\left(\frac{n-1}{2}\right)[2\times 4+(n-1-1\left)2\right]$

$=3+\left(\frac{n-1}{2}\right)[8+(n-2\left)2\right]$

$=3+\left(\frac{n-1}{2}\right)(2n+4)$

$=3+(n-1)(n+2)$

$=3+(n^2+n-2)$

$=n^2+n+1$

$\therefore S={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1$

$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$ [by using ${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$]

$=n\left[\frac{(n+1)(2n+1)+3(n+1)6}{6}\right]$

$=n\left[\frac{{2n}^2+3n+1+3n+3+6}{6}\right]$

$=n\left[\frac{{2n}^2+6n+10}{6}\right]$

$=\frac{n}{3}(n^2+3n+5)$